# The Unapologetic Mathematician

## Improper Integrals II

I’ve got one last thing to wrap up in my coverage of integration. Well, I never talked about Brown v. Board of Ed., but most curricula these days don’t cover the technique “by court order” anymore.

Did I get the title of the post wrong? No. It turns out that I covered half of this topic two years ago as I prepared to push into infinite series. Back then, I dealt with what happened when we wanted to integrate over an infinite interval. Then I defined an infinite series to be what happened when we used a particular integrator in our Riemann-Stieltjes integral. This could also be useful in setting up multiple integrals over $n$-dimensional intervals that extend infinitely far in some direction.

But there are other places that so-called “improper integrals” come up, like the example I worked through the other day. In this case, the fundamental theorem of calculus runs into trouble at the endpoints of our interval. Indeed, we ask for an antiderivative on a closed interval, but to get the endpoints we need to have the antiderivative defined on some slightly larger open interval, and it just isn’t.

So here’s what we do: let $f$ be defined on $(a,b]$ and integrable with respect to some integrator $\alpha$ over the interval $[x,b]$ for all $x\in(a,b]$. Then we can define each integral

$\displaystyle\int\limits_x^bf\,d\alpha$

Just as before, we define the improper integral to be the one-sided limit

$\displaystyle\int\limits_{a^+}^bf\,d\alpha=\lim\limits_{x\rightarrow a^+}\int\limits_x^bf\,d\alpha$

If this limit exists, we say that the improper integral converges. Otherwise we say it diverges. Similarly we can define the improper integral

$\displaystyle\int\limits_a^{b^-}f\,d\alpha=\lim\limits_{x\rightarrow b^-}\int\limits_a^xf\,d\alpha$

taking the limit as $x$ approaches $b$ from the left.

Just as for the first kind of improper integral, we have analogues of the direct comparison and limit comparison tests, and of the idea of absolute convergence. Each of these is exactly the same as before, replacing limits as $x$ approaches $\infty$ with limits as $x$ approaches some finite point from the left or right.

We can also combine improper integrals like we did before to integrate over the whole real line. For example, we could define

$\displaystyle\int\limits_{a^+}^{b^-}f\,d\alpha=\lim\limits_{\substack{x\rightarrow a^+\\y\rightarrow b^-}}\int\limits_x^yf\,d\alpha$

or

$\displaystyle\int\limits_{a^+}^\infty f\,d\alpha=\lim\limits_{\substack{x\rightarrow a^+\\y\rightarrow\infty}}\int\limits_x^yf\,d\alpha$

As before, we must take these limits separately. Indeed, now we can even say more about what can go wrong, because these are examples of multivariable limits. We cannot take the limit as $x$ and $y$ together approach their limiting points along any particular path, but must consider them approaching along all paths.

In practice, it’s pretty clear what needs to be done, and when. If we have trouble evaluating an antiderivative at one endpoint or another, we replace the evaluation with an appropriate limit.

## Differentiating Partial Integrals

A neat little variation on differentiating an integral from last time combines it with the fundamental theorem of calculus. It’s especially interesting in the context of evaluating iterated integrals for irregular regions where the limits of integration may depend on other variables.

Let $f:R\rightarrow\mathbb{R}$ is a continuous function on the rectangle $[a,b]\times[c,d]$, and that $D_2f$ is also continuous on $R$. Also, let $p:[c,d]\rightarrow[a,b]$ and $q:[c,d]\rightarrow[a,b]$ be two differentiable functions on $[c,d]$ with images in $[a,b]$. Define the function

$\displaystyle F(y)=\int\limits_{p(y)}^{q(y)}f(x,y)\,dx$

Then the derivative $F'(y)$ exists and has the value

$\displaystyle F'(y)=\int\limits_{p(y)}^{q(y)}\left[D_2f\right](x,y)\,dx+f(q(y),y)q'(y)-f(p(y),y)p'(y)$

In fact, if we forget letting $f$ depend on $y$ at all, this is the source of some of my favorite questions on first-semester calculus finals.

Anyway, we define another function for the moment

$\displaystyle G(x^1,x^2,x^3)=\int\limits_{x^1}^{x^2}f(t,x^3)\,dt$ for $x^1$ and $x^2$ in $[a,b]$ and $x^3$ in $[c,d]$. Then $F(y)=G(p(y),q(y),y)$.

The fundamental theorem of calculus tells us the first two partial derivatives of $G$ immediately, and for the third we can differentiate under the integral sign:

\displaystyle\begin{aligned}\frac{\partial G}{\partial x^1}&=-f(x^1,x^3)\\\frac{\partial G}{\partial x^2}&=f(x^2,x^3)\\\frac{\partial G}{\partial x^3}&=\int\limits_{x^1}^{x^2}\left[D_2f\right](t,x^3)\,dt\end{aligned}

Then we can use the chain rule:

\displaystyle\begin{aligned}\frac{d}{dy}F(y)&=\frac{d}{dy}G(p(y),q(y),y)\\&=\frac{\partial G}{\partial x^1}\frac{dp}{dy}+\frac{\partial G}{\partial x^2}\frac{dq}{dy}+\frac{\partial G}{\partial x^3}\frac{dy}{dy}\\&=-f(p(y),y)p'(y)+f(q(y),y)q'(y)+\int\limits_{p(y)}^{q(y)}\left[D_2f\right](x,y)\,dx\end{aligned}

January 14, 2010 Posted by | Analysis, Calculus | 3 Comments

## Differentiation Under the Integral Sign

Another question about “partial integrals” is when we can interchange integral and differential operations. Just like we found for continuity of partial integrals, this involves interchanging two limit processes.

Specifically, let’s again assume that $f:R\rightarrow\mathbb{R}$ is a function on the rectangle $R=[a,b]\times[c,d]$, and that $\alpha$ is of bounded variation on $[a,b]$ and that the integral

$\displaystyle F(y)=\int\limits_a^bf(x,y)\,d\alpha(x)$

exists for every $y\in[c,d]$. Further, assume that the partial derivative $\left[D_2f\right](x,y)$ exists and is continuous throughout $R$. Then the derivative of $F$ exists and is given by

$\displaystyle\frac{d}{dy}F(y)=\frac{d}{dy}\int\limits_a^bf(x,y)\,d\alpha(x)=\int\limits_a^b\frac{\partial}{\partial y}f(x,y)\,d\alpha(x)$

Similar results hold where $x$ and $y$ are vector values, and where derivatives in terms of $y$ are replaced outside the integral by partial derivatives in terms of its components.

So, if $y_0\in(c,d)$ and $y\neq y_0$ we can calculate the difference quotient:

\displaystyle\begin{aligned}\frac{F(y)-F(y_0)}{y-y_0}&=\int\limits_a^b\frac{f(x,y)-f(x,y_0)}{y-y_0}\,d\alpha(x)\\&=\int\limits_a^b\left[D_2f\right](x,\xi)\,d\alpha(x)\end{aligned}

where $\xi$ is some number between $y_0$ and $y$ that exists by the differential mean value theorem. Now to find the derivative, we take the limit of the difference quotient

\displaystyle\begin{aligned}\frac{d}{dy}F(y_0)&=\lim\limits_{y\rightarrow y_0}\frac{F(y)-F(y_0)}{y-y_0}\\&=\lim\limits_{y\rightarrow y_0}\int\limits_a^b\left[D_2f\right](x,\xi)\,d\alpha(x)\\&=\int\limits_a^b\lim\limits_{y\rightarrow y_0}\left[D_2f\right](x,\xi)\,d\alpha(x)\end{aligned}

as we take $y$ to approach $y_0$, the number $\xi$ gets squeezed towards $y_0$ as well. Since we assumed $\left[D_2f\right](x,y)$ to be continuous, the limit in the integrand will equal $\left[D_2f\right](x,y_0)$, as asserted.

January 13, 2010 Posted by | Analysis, Calculus | 13 Comments

## Continuity of Partial Integrals

There are some remaining topics to clean up in the theory of the Riemann-Stieltjes integral. First up is a question that seems natural from the perspective of iterated integrals: what can be said about the continuity of the inner integrals?

More explicitly, let $f:R\rightarrow\mathbb{R}$ be a continuous function on the rectangle $R=[a,b]\times[c,d]$, and let $\alpha$ be of bounded variation on $[a,b]$. Define the function $F$ on $[c,d]$ by

$\displaystyle F(y)=\int\limits_a^bf(x,y)\,d\alpha(x)$

Then $F$ is continuous on $[c,d]$. Similar statements can be made about other “partial integrals”, where $x$ and $y$ are each vector variables and we use $d(x^1,\dots,x^k)$ in place of the Stieltjes integrator $d\alpha(x)$.

Specifically, this is a statement about interchanging limit operations. The Riemann-Stieltjes integral involves taking the limit over the collection of tagged partitions of $[a,b]$, while to ask if $F$ is continuous asks whether

$\displaystyle\lim\limits_{y\rightarrow y_0}\int\limits_a^bf(x,y)\,d\alpha(x)=\int\limits_a^b\lim\limits_{y\rightarrow y_0}f(x,y)\,d\alpha(x)=\int\limits_a^bf(x,y_0)\,d\alpha(x)$

As we knew back when we originally discussed integrators of bounded variation, we can write our integrator as the difference of two increasing functions. It’s no loss of generality, then, to assume that $\alpha$ is increasing. We also remember that the Heine-Cantor theorem tells us that since $R$ is compact, $f$ is actually uniformly continuous.

Uniform continuity tells us that for every $\epsilon>0$ there is a $\delta>0$ (depending only on $\epsilon$ so that for every pair of points $(x,y)$ and $(x',y')$, with $\lvert(x,y)-(x',y')\rvert<\delta$ we have $\lvert f(x,y)-f(x',y')\rvert<\epsilon$.

So now let’s take two points $y$ and $y'$ with $\lvert y-y'\rvert<\delta$ and consider the difference

\displaystyle\begin{aligned}\lvert F(y)-F(y')\rvert&=\left\lvert\int\limits_a^bf(x,y)\,d\alpha(x)-\int\limits_a^bf(x,y')\,d\alpha(x)\right\rvert\\&=\left\lvert\int\limits_a^bf(x,y)-f(x,y')\,d\alpha(x)\right\rvert\\&\leq\int\limits_a^b\lvert f(x,y)-f(x,y')\rvert\,d\alpha(x)\\&\leq\epsilon\left(\alpha(b)-\alpha(a)\right)\end{aligned}

where we’ve used the integral mean value theorem. Clearly by choosing the right $\epsilon$ we can find a $\delta$ to make the right hand side as small as we want, proving the continuity of $F$.

January 12, 2010 Posted by | Analysis, Calculus | 1 Comment

## The Geometric Interpretation of the Jacobian Determinant

We first defined the Jacobian determinant as measuring the factor by which a transformation scales infinitesimal pieces of $n$-dimensional volume. Now, with the change of variables formula and the mean value theorem in hand, we can pull out a macroscopic result.

If $g:S\rightarrow\mathbb{R}^n$ is injective and continuously differentiable on an open region $S\subseteq\mathbb{R}^n$, and $X$ is a compact, connected, Jordan measurable subset of $S$, then we have

$\displaystyle c(Y)=\lvert J_h(x_0)\rvert c(X)$

for some $x_0\in X$, where $Y=g(X)$.

Simply take $f(y)=1$ in the change of variables formula

$\displaystyle c(Y)=\int\limits_Y\,dy=\int\limits_X\lvert J_g(x)\rvert\,dx$

The mean value theorem now tells us that

$\displaystyle c(Y)=\int\limits_X\lvert J_g(x)\rvert\,dx=\lambda\int\limits_X\,dx=\lambda c(X)$

for some $\lambda$ between the maximum and minimum of $\lvert J_g(x)\rvert$ on $X$. But then since $X$ is connected, we know that there is some $x_0\in X$ so that $\lambda=\lvert J_g(x_0)\rvert$, as we asserted.

January 8, 2010 Posted by | Analysis, Calculus | 10 Comments

## Change of Variables in Multiple Integrals III

Today we finish up the proof of the change of variables formula for multiple integrals:

$\displaystyle\int\limits_Xf(x^1,\dots,x^n)\,d(x^1,\dots,x^n)=\int\limits_{g^{-1}(X)}f(g(u^1,\dots,u^n))\left\lvert\frac{\partial(x^1,\dots,x^n)}{\partial(u^1,\dots,u^n)}\right\rvert\,d(u^1,\dots,u^n)$

So far, we’ve shown that we can chop $X$ up into a collection of nonoverlapping regions $T_{(k)}$ and $g^{-1}(X)$ into their preimages $A_{(k)}=g^{-1}(T_{(k)})$. Further, within each $A_{(k)}$ we can factor $g(u)=\theta_{(k)}(\phi_{(k)}(u))$, where $\phi_{(k)}$ fixes each component of $u$ except the last, and $\theta_{(k)}$ fixes that one. If we can show the formula holds for each such region, then it will hold for arbitrary (compact, Jordan measurable) $X$.

From here we’ll just drop the subscripts to simplify our notation, since we’re just concerned with one of these regions at a time. We’ve got $T$ and its preimage $A=g^{-1}(T)$. We’ll also define $B=\phi(A)$, so that $T=\theta(B)$. For each real $\xi$ we define

\displaystyle\begin{aligned}T(\xi)&=\{(x^1,\dots,x^{n-1}\vert(x^1,\dots,x^{n-1},\xi)\in T\}\\B(\xi)&=\{(t^1,\dots,t^{n-1})\vert(t^1,\dots,t^{n-1},\xi)\in B\}\end{aligned}

Then $(T(\xi),\xi)=\theta(B(\xi),\xi)$, since $\theta$ preserves the last component of the vector. We also define

\displaystyle\begin{aligned}c&=\inf\{\phi^n(u)\vert u\in A\}\\d&=\sup\{\phi^n(u)\vert u\in A\}\end{aligned}

The lowest and highest points in $T$ along the $n$th coordinate direction. Now we can again define $F(x)=f(x)\chi_T(x)$ and set up the iterated integral

$\displaystyle\int\limits_TF(x)dx=\int\limits_c^d\int\limits_{T(x^n)}F(x^1,\dots,x^{n-1},x^n)\,d(x^1,\dots,x^{n-1})\,dx^n$

We can apply the inductive hypothesis to the inner integral using $x=\theta(t)$, which only involves the first $n-1$ coordinates anyway. If we also rename $x^n$ to $t^n$, this gives

$\displaystyle\int\limits_TF(x)dx=\int\limits_c^d\int\limits_{B(t^n)}F(\theta(t^1,\dots,t^{n-1},t^n))\lvert J_\theta(t)\rvert\,d(t^1,\dots,t^{n-1})\,dt^n$

Which effectively integrates as $t$ runs over $B=\phi(A)$. But now we see that $B(t^n)$ lies within the projection $A_n$ of $A$, as we defined when we first discussed iterated integrals. We want to swap the order of integration here, so we have to rewrite the limits. To this end, we write $A=[a^1,b^1]\times\dots\times[a^n,b^n]$, $A_n=[a^1,b^1]\times\dots\times[a^{n-1},b^{n-1}]$, and define

$B^*(u^1,\dots,u^{n-1})=\{\phi^n(u^1,\dots,u^{n-1},u^n)\vert a^n\leq u^n\leq b^n\}$

which runs over the part of $B$ above some fixed point in $A_n$. Then we can reverse the order of integration to write

$\displaystyle\int\limits_TF(x)dx=\int\limits_{A_n}\int\limits_{B^*(u^1,\dots,u^{n-1})}F(\theta(t^1,\dots,t^{n-1},t^n))\lvert J_\theta(t)\rvert\,dt^n\,d(t^1,\dots,t^{n-1})$

Now we can perform the one-dimensional change of variables on the inner integral and swap out the variables $u^1=t^1$ through $u^{n-1}=t^{n-1}$ to write

$\displaystyle\int\limits_TF(x)dx=\int\limits_{A_n}\int\limits_{a^n}^{b^n}F(\theta(\phi(u)))\lvert J_\theta(\phi(u))\rvert\lvert J_\phi(u)\rvert\,du^n\,d(u^1,\dots,u^{n-1})$

But now we recognize the product of the two Jacobian determinants as the Jacobian of the composition:

$\displaystyle J_\theta(\phi(u))J_\phi(u)=J_{\theta\circ\phi}(u)$

and so we can recombine the iterated integral into the $n$-dimensional integral

$\displaystyle\int\limits_TF(x)dx=\int\limits_AF([\theta\circ\phi](u))\lvert J_{\theta\circ\phi}(u)\rvert\,du$

Finally, since $g=\theta\circ\phi$ we can replace the composition. We can also replace $F$ by $f$ since there’s no chance anymore for any evaluation of $f$ to go outside $T$. We’re left with

$\displaystyle\int\limits_Tf(x)dx=\int\limits_{g^{-1}(T)}f(g(u))\lvert J_g(u)\rvert\,du$

Essentially, what we’ve shown is that we can always arrange to first change $n-1$ variables (which we handle by induction) and then by the last variable. The overall scaling factor is the $n-1$-dimensional scaling factor from the first transformation times the one-dimensional scaling factor from the second, and this works because of how the Jacobian works with compositions.

This establishes the change of variables formula for regions within which we can write $g$ as the composition of two functions, one of which fixes all but the last coordinate, and the other of which fixes that one. Since we established that we can always cut up a compact, Jordan measurable set $X$ into a finite number of such pieces, this establishes the change of variables formula in general.

January 7, 2010 Posted by | Analysis, Calculus | 4 Comments

## Change of Variables in Multiple Integrals II

Okay, let’s get to actually proving the change of variables formula for multiple integrals. To be explicit:

Let $g=(g^1,\dots,g^n)$ be a continuously differentiable function defined on an open region $X\subseteq\mathbb{R}^n$. Further, assume that $g$ is injective and that the Jacobian determinant $J_g$ is everywhere nonzero on $S$. The inverse function theorem tells us that we can define a continuously differentiable inverse $g^{-1}$ on all of the image $g(S)$.

Further, let $X$ be a Jordan measurable subset of $g(S)$, and let $f$ be defined and continuous on $X$. Then we have the change of variables formula

$\displaystyle\int\limits_Xf(x^1,\dots,x^n)\,d(x^1,\dots,x^n)=\int\limits_{g^{-1}(X)}f(g(u^1,\dots,u^n))\left\lvert\frac{\partial(x^1,\dots,x^n)}{\partial(u^1,\dots,u^n)}\right\rvert\,d(u^1,\dots,u^n)$

We will proceed by induction on the dimension $n$. For $n=1$, this is exactly the one-dimensional change of variables formula, which we already know to be true. And so we’ll assume that it holds in all $n-1$-dimensional cases, and prove that it then holds for $n$-dimensional cases as well.

Since the Jacobian determinant $J_g(v)$ is nonzero, we cannot have $\left[D_ng^k\right](v)=0$ for all $k$; at least one of the components must have a nonzero partial derivative with respect to $v^n$ at any given point $v\in S$. Let’s say, to be definite, that $\left[D_ng^n\right](v)\neq0$. We will now (locally) factor $g$ into the composite of two functions $\theta$ and $\phi$, which will each have their own useful properties. First, we will define

$\displaystyle\phi(u)=(u^1,\dots,u^{n-1},g^n(u))$

This is clearly continuously differentiable, and it’s even injective on some neighborhood of $v$, by our assumption that $\left[D_ng^n\right](v)\neq0$. Further, the Jacobian determinant $J_\phi$ is exactly the partial derivative $D_ng^n$, and so the inverse function theorem tells us that in some neighborhood of $v$ we have a local inverse $\psi$, with $\psi(\phi(u))=u$ in some neighborhood of $v$. We can now define

$\displaystyle\theta^k(t)=g^k(t^1,\dots,t^{n-1},\psi^n(t))$

for $1\leq k\leq n-1$, and define

$\displaystyle\theta^n(t)=t^n$

Then for each $u$ in a small enough neighborhood of $v$ we have $\theta(\phi(u))=g(u)$. The first function $\phi$ leaves all components of $u$ fixed except for $u^n$, while the second function $\theta$ leaves $u^n$ fixed. Of course, if we used a different partial derivative $D_ng^k$, we could do the same thing, replacing $u^n$ instead with $D_ng^k$ in $\phi$, and so on.

Now if $X$ is a Jordan measurable compact subset of $g(S)$, then its inverse image $g^{-1}(X)$ will also be compact since $g^{-1}$ is continuous. For every point in $g^{-1}(X)\subseteq S$, we can find some neighborhood — which we can take to be an $n$-dimensional interval — and a factorization of $g$ into two functions as above. As we move around, these neighborhoods form an open cover of $g^{-1}(X)$. And since $g^{-1}(X)$ is compact, we can take an open subcover.

That is, we can cover $g^{-1}(X)$ by a finite collection of open intervals $A_{(k)}$, and within each one we can write $g(u)=\theta_{(k)}(\phi_{(k)}(u))$, where the function $\phi_{(k)}$ leaves all the components of $u$ fixed except for the last, while $\theta_{(k)}$ leaves that last one fixed. By subdividing these intervals, we can assume that they’re nonoverlapping. Of course, if we subdivided into open sets we’d miss the shared boundary between two subintervals. So we’ll include that boundary in each subinterval and still have $p$ nonoverlapping intervals $A_{(k)}$.

Then we can define $T_{(k)}=g(A_{(k)})$. Since $g$ is injective, these regions will also be nonoverlapping. And they’ll cover $X$, just as $A_{(k)}$ covered $g^{-1}(X)$. So we can define $F(x)=f(x)\chi_X(x)$ and write

$\displaystyle\int\limits_Xf(x^1,\dots,x^n)\,d(x^1,\dots,x^n)=\sum\limits_{k=1}^p\int\limits_{T_{(k)}}F(x^1,\dots,x^n)\,d(x^1,\dots,x^n)$

Our proof will thus be complete if we can show that the change of variables formula holds for these regions, and we will pick up with this final step next time.

January 6, 2010 Posted by | Analysis, Calculus | 3 Comments

## Change of Variables in Multiple Integrals I

In the one-variable Riemann and Riemann-Stieltjes integrals, we had a “change of variables” formula. This let us replace our variable of integration by a function of a new variable, and we got the same answer. This was useful because the form of the resulting integrand might have been simpler to work with in terms of using the fundamental theorem of calculus.

In multiple variables we’ll have a similar formula, but it will have an additional use. Not only might it be used to simplify the integrand, but it might simplify the region of integration itself! Of course, there might also be a trade-off between these two considerations, as many students in multivariable calculus classes might remember. A substitution which simplifies the region of integration might make antidifferentiating the integrand (in any of the resulting variables) impractical, while another substitution which simplifies the integrand might make the region a nightmare to work with.

The formula in one variable looked something like this:

$\displaystyle\int\limits_a^bf(x)\,dx=\int\limits_c^df(g(u))g'(u)\,du$

where $x=g(u)$ (along with the induced transformation $dx=g'(u)\,du$) is a continuously differentiable function on $u\in[c,d]$ with $a=g(c)$ and $b=g(d)$. Notice that $g(u)$ could extend out beyond $[a,b]$, but if it went above $b$ it would have to come back down again, covering the same region twice with opposite signs. This is related to the signed volumes we talked about, where (in one dimension) an interval can be traversed (integrated over) from left to right or from right to left.

The picture gets a little simpler when we assume that $g$ is strictly monotonic. That is, either $g$ is strictly increasing, $g'(u)>0$, and $g([c,d])=[a,b]$; or $g$ is strictly decreasing, $g'(u)<0$, and $g([c,d])=[b,a]$ (traversing in the opposite direction). In the first case, we can write our change of variables relation as

$\displaystyle\int\limits_{[a,b]}f(x)\,dx=\int\limits_{g^{-1}[a,b]}f(g(u))g'(u)\,du$

while in the second case, reversing the direction of integration entails adding a negative sign.

$\displaystyle\int\limits_{[a,b]}f(x)\,dx=-\int\limits_{g^{-1}[a,b]}f(g(u))g'(u)\,du$

but in this case, the derivative $g'(u)$ is strictly negative. We can combine it with this new sign, and rather elegantly write both cases as

$\displaystyle\int\limits_{[a,b]}f(x)\,dx=\int\limits_{g^{-1}[a,b]}f(g(u))\lvert g'(u)\rvert\,du$

In all of these cases, we know that the inverse function exists because of the inverse function theorem. Here the Jacobian determinant is simply the derivative $g(x)$, which we’re assuming is everywhere nonzero.

In essence, the idea was that $\lvert g'(u)\rvert$ measures the factor by which $g$ stretches intervals near $u$. That is, the tiny bit of one-dimensional volume $du$ gets stretched into the tiny bit of (unsigned) one-dimensional volume $dx=\lvert g'(u)\rvert\,du$. And this works because at a very small scale, little changes in $u$ transform almost linearly.

So in higher-dimensional spaces, we will assume that $g$ transforms small enough changes in $u$ almost linearly — $g$ is differentiable — and that the Jacobian determinant $J_g(u)$ is everywhere nonzero, so we can invert the transformation. This gives us hope that we can write something like

$\displaystyle\int\limits_{S}f(x)\,dx=\int\limits_{g^{-1}(S)}f(g(u))\,dx$

Since $g$ is invertible, integrating as $u$ ranges over $g^{-1}(S)$ is the same as letting $x=g(u)$ range over $S$, so the region of integration lines up, as does the integrand. All that’s left is to figure out how we should replace $dx$.

Now this $dx$ is not the differential of a variable $x$. When it shows up in a multiple integral, it’s a tiny little bit of $n$-dimensional volume. And we measure the scaling of $n$-dimensional volumes with the Jacobian determinant! The same sign considerations as before tell us that either the Jacobian determinant is always positive or always negative, and in either case we can write

$\displaystyle\int\limits_{S}f(x)\,dx=\int\limits_{g^{-1}(S)}f(g(u))\lvert J_g(u)\rvert\,du$

or, using our more Leibniz-style notation

$\displaystyle\int\limits_{S}f(x^1,\dots,x^n)\,d(x^1,\dots,x^n)=\int\limits_{g^{-1}(S)}f(g(u^1,\dots,u^n))\left\lvert\frac{\partial(x^1,\dots,x^n)}{\partial(u^1,\dots,u^n)}\right\rvert\,d(u^1,\dots,u^n)$

We will start proving this formula next time.

January 5, 2010 Posted by | Analysis, Calculus | 5 Comments

## An Example of an Iterated Integral

My description of how to evaluate a multiple integral over some region other than an $n$-dimensional interval by using iterated integrals might not have been the clearest, so I’m hoping an example will help illustrate what I mean. Let’s calculate the Jordan content of a three-dimensional sphere of radius $a$, centered at the origin. This will also help cement the fact that Jordan content is closely related to what we mean by “volume”.

So, we have our sphere

$\displaystyle S=\left\{(x,y,z)\in\mathbb{R}^3\big\vert x^2+y^2+z^2\leq a^2\right\}$

We can put it inside the interval $R=[-a,a]\times[-a,a]\times[-a,a]$, so we know that the Jordan content is

$\displaystyle c(S)=\int\limits_S\,d(x,y,z)=\int\limits_R\chi_S(x,y,z)\,d(x,y,z)$

Now we want to peel off an integral from the inside. Let’s first integrate over the variable $z$. Projecting the sphere onto the plane $z=0$ we’re left with the circle

$\displaystyle S_z=\left\{(x,y)\in\mathbb{R}^2\big\vert x^2+y^2\leq a^2\right\}$

and we need to write $S$ as the region between the graphs of two functions on this projection. And indeed, we can write

$\displaystyle S=\left\{(x,y,z)\in\mathbb{R}^3\big\vert(x,y)\in S_z,-\sqrt{a^2-x^2-y^2}\leq z\leq\sqrt{a^2-x^2-y^2}\right\}$

Thus we can write

$\displaystyle\int\limits_S\,d(x,y,z)=\int\limits_{S_z}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dz\,d(x,y)$

Next we’ll integrate with respect to $y$. Projecting $S_z$ onto the line $y=0$ we find $S_{zy}=[-a,a]$. And then we find

$\displaystyle S_z=\left\{(x,y)\in\mathbb{R}^2\big\vert-a\leq x\leq a,-\sqrt{a^2-x^2}\leq y\leq\sqrt{a^2-x^2}\right\}$

which lets us peel off another integral from the inside

$\displaystyle\int\limits_S\,d(x,y,z)=\int\limits_{[-a,a]}\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dz\,dy\,dx=\int\limits_{-a}^a\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dz\,dy\,dx$

Finally, we can start evaluating these integrals. The innermost integral goes immediately, using the fundamental theorem of calculus. An antiderivative of $1$ with respect to $z$ is $z$, and so we find

\displaystyle\begin{aligned}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dz&=z\bigg\vert_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\\&=\sqrt{a^2-x^2-y^2}-(-\sqrt{a^2-x^2-y^2})\\&=2\sqrt{a^2-x^2-y^2}\end{aligned}

So we put this into our integral, already in progress:

$\displaystyle\int\limits_S\,d(x,y,z)=\int\limits_{-a}^a\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dz\,dy\,dx=\int\limits_{-a}^a\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}2\sqrt{a^2-x^2-y^2}\,dy\,dx$

Here’s where things start to get difficult. I’ll just tell you (and you can verify) that an antiderivative of $2\sqrt{a^2-x^2-y^2}$ with respect to $y$ is

$\displaystyle y\sqrt{a^2-x^2-y^2}+(a^2-x^2)\arctan\left(\frac{y}{\sqrt{a^2-x^2-y^2}}\right)$

So the fundamental theorem of calculus tells us

$\displaystyle\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}2\sqrt{a^2-x^2-y^2}\,dy=(a^2-x^2)\left(\arctan\left(\frac{\sqrt{a^2-x^2}}{0^+}\right)-\arctan\left(\frac{-\sqrt{a^2-x^2}}{0^+}\right)\right)$

Technically, this is an improper integral (which we haven’t really discussed), so we need to take some limits. The denominators $0^+$ are limits as some dummy variable approaches zero from above. Thus we continue evaluating our integral

$\displaystyle\int\limits_S\,d(x,y,z)=\int\limits_{-a}^a\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}2\sqrt{a^2-x^2-y^2}\,dy\,dx=\int\limits_{-a}^a\pi(a^2-x^2)\,dx$

This one is easy: an antiderivative of $a^2-x^2$ is $a^2x-\frac{x^3}{3}$, and so

\displaystyle\begin{aligned}\int\limits_S\,d(x,y,z)&=\int\limits_{-a}^a\pi(a^2-x^2)\,dx\\&=\pi\left(a^3-\frac{a^3}{3}\right)-\pi\left(-a^3-\frac{(-a)^3}{3}\right)\\&=\frac{4}{3}\pi a^3\end{aligned}

And this is exactly the well-known formula for the volume of a sphere of radius $a$. There are easier ways to get at this formula, of course, but this one manages to illustrate the technique of iterated integrals with variable limits of integration on the inner integrals.

January 4, 2010 Posted by | Analysis, Calculus | 3 Comments